proof of the sheldon conjecture - dartmouth collegecarlp/sheldon022119.pdffor part (6), we direct...

Mathematical Assoc. of America American Mathematical Monthly 121:1 February 21, 2019 12:47 p.m. ”sheldon 2.21.2019”.tex page 1

Proof of the Sheldon ConjectureCarl Pomerance and Chris Spicer

Abstract. In [3], the authors introduce the concept of a Sheldon prime, based on a conversationbetween several characters in the CBS television situation comedy The Big Bang Theory. Theauthors of [3] leave open the question of whether 73 is the unique Sheldon prime. This paperanswers this question in the affirmative.

1. INTRODUCTION. A Sheldon prime was first defined in [3] as an homage toSheldon Cooper, a fictional theoretical physicist, see Figure 1, on the television showThe Big Bang Theory, who claimed 73 is the best number because it has some seem-ingly unusual properties. First note that not only is 73 a prime number, its index inthe sequence of primes is the product of its digits, namely 21: it is the 21st prime. Inaddition, reversing the digits of 73, we obtain the prime 37, which is the 12th prime,and 12 is the reverse of 21.

We give a more formal definition. For a positive integer n, let pn denote the nthprime number. We say pn has the product property if the product of its base-10 digitsis precisely n. For any positive integer x, we define rev(x) to be the integer whosesequence of base-10 digits is the reverse of the digits of x. For example, rev(1234) =4321 and rev(310) = 13. We say pn satisfies the mirror property if rev(pn) = prev(n).

Definition. The prime pn is a Sheldon prime if it satisfies both the product propertyand the mirror property.

In [3], the “Sheldon Conjecture” was posed that 73 is the only Sheldon prime. InSection 5 we prove the following result.

Theorem 1. The Sheldon conjecture holds: 73 is the unique Sheldon prime.

2. THE PRIME NUMBER THEOREM AND SHELDON PRIMES. Let π(x) de-note the number of prime numbers in the interval [2, x]. Looking at tables of primesit appears that they tend to thin out, becoming rarer as one looks at larger numbers.This can be expressed rigorously by the claim that lim

x→∞π(x)/x = 0. In fact, more is

true: we know the rate at which the ratio π(x)/x tends to 0. This is the prime numbertheorem:

limx→∞

π(x)

x/ log x= 1,

where “log” is the natural logarithm function. This theorem was first proved in 1896independently by Hadamard and de la Vallee Poussin, following a general plan laidout by Riemann about 40 years earlier (the same paper where he first enunciated thenow famous Riemann hypothesis).

We actually know that π(x) is slightly larger than x/ log x for large values of x;in fact there is a secondary term x/(log x)2, a positive tertiary term, and so on. Thephrase “large values of x” can be made numerically explicit: A result of Rosser andSchoenfeld [7, (3.5)] is that

π(x) >x

log xfor all x ≥ 17. (1)

January 2014] PROOF OF THE SHELDON CONJECTURE 1


This beautiful inequality immediately allows us to prove that no Sheldon prime ex-ceeds 1045, and in fact, we only need the product property to show this.

Figure 1. Sheldon always knew 73 was the best.PHOTO CREDIT: Michael Yarish/©2019 Warner Bros. Entertainment Inc.

Proposition 2. If pn has the product property, then pn < 1045.

Proof. Say pn has k digits with the leading digit a. Then n, which is equal to theproduct of the digits of pn, is at most a×9k−1. Using (1), for pn ≥ 17, we have

n = π(pn) >pn

log pn.

But pn ≥ a×10k−1 since pn is k digits long. Thus if pn has the product property, thenthe following inequality must be satisfied:

a×9k−1 > a×10k−1

log(a×10k−1),

which implies that

log a+ log(10k−1) >

(10

9

)k−1

. (2)

Since the left side grows linearly in k and the right side grows exponentially, it is clearthat (2) fails for all large values of k. Further, if (2) fails for a = 9, then it also failsfor smaller values of a. A small computation and mathematical induction allow us tosee that (2) fails for all k ≥ 46.

Finer estimates than (1) exist in the literature, some of which are referenced belowin Section 4. However, they do not afford much of an improvement in Proposition 2.

In [3], the authors show that the primes p7 = 17, p21 = 73, and p181,440 =2,475,989 each satisfy the product property. This leads us to the following conjecture.

2 © THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121


Conjecture 3. The only primes with the product property are

p7 = 17, p21 = 73, and p181,440 = 2,475,989.

We have exhaustively searched for primes pn with the product property for all n ≤1010 (using the built-in Mathematica function that gives the nth prime), and foundonly the 3 examples listed above. It is certainly possible to extend this search, but itseems computationally challenging to cover all of the territory up to 1045.

An example of a challenging number to analyze is

n = 276,468,770,930,688 = 21731672.

It is not impossible, but difficult, to compute pn. Short of this, if only we could approx-imate pn we might be able to determine its most significant digits, which may allowus to rule it out. As discussed in Section 4 below, this approximation is afforded by theinverse function of the logarithmic integral function, namely li−1(n). Definitions willbe forthcoming, but for now note that Lemma 7 shows that

9,897,979,324,865,422 < pn < 9,897,979,533,554,693. (3)

We deduce that the top 7 digits of pn are 9, 8, 9, 7, 9, 7, 9, and the 8th digit must be a3, 4, or 5. The product of the first 7 digits is 2,571,912, and the quotient after dividingthis into n is 107,495,424 = 21438. If pn were to satisfy the product property, we seethat the remaining 9 digits in pn must consist of four 9’s, four 8’s, and one 4. Thus, wemay assume the 8th digit of pn is 4, and the last digit is 9. There are still 35 possibilitiesfor the placement of the remaining digits. Although we may hope each would result ina composite number, that is not the case. For example, we have the candidates

9,897,979,489,888,999,

9,897,979,489,989,889,

9,897,979,489,998,889,

9,897,979,498,889,899.

Each of the above is prime, the product of their digits is n, and the only thing in doubtis their indices in the sequence of primes. These indices are all near n, but there arestill many possibilities. It is certainly a tractable problem to find these indices, but itseems there will be many similar and much harder challenges as one searches higher.

To prove our theorem that 73 is the only Sheldon prime we will make use of themirror property in addition to the product property. For example, the reverses of the 4primes above are all composite, so they are instantly ruled out as Sheldon primes.

Though the bound 1045 may seem daunting, we see at least that the search forSheldon primes is finite. Our basic strategy is to use numerically explicit versions ofthe prime number theorem, similar to, but finer than, (1) to give us some of the leadingand trailing digits of candidate primes, and use these to, we hope, eliminate them.Further, our search is not over all primes to 1045 but over integers n with pn < 1045.

But first we need to assemble our weapons for the attack!

3. PROPERTIES OF SHELDON PRIMES. Because a Sheldon prime must satisfyboth the product property and the mirror property (described in the Introduction), thereare a few simple tests one can apply to candidates based on properties of Sheldonprimes.



Proposition 4. If pn is a Sheldon prime and n > 1010, then

1. n is 7-smooth (meaning that no prime dividing n exceeds 7);2. the leading digit of pn must be in {1, 3, 7, 9};3. the number of digits of prev(n) must equal the number of digits of pn;4. 54 - n;5. if pn > 1019, then 53 - n;6. 100 - n;7. pn cannot have a digit 0, and cannot have a digit 1 except possibly for the

leading digit;8. the leading digit of prev(n) must be in {3, 7, 9}.

Proof. Part (1) is immediate from the definition of the product property. Parts (2) and(3) are clear since rev(pn) must be prime, and primes beyond single digits must endin 1, 3, 7, or 9. Noting that each factor 5 in n must come from a digit 5 in pn, one canprove part (4) using the same method as the proof of Proposition 2. In particular,

a×54×9k−5 < a×10k−1

log(a×10k−1)

for a = 1, 3, 7, or 9 and k ≥ 5. One can similarly derive part (5).For part (6), we direct the reader to [3], where a detailed proof is given. The idea

is that if 100 | n, then rev(n) < 110n, while rev(pn) has the same number of digits as

pn. Prime number inequalities, such as (6), complete the proof.It is obvious that no prime having the product property can have a digit 0. For the

second part of (7), suppose that pn has a digit 1 after the leading digit. But

a×9k−2 < a×10k−1

log(a×10k−1)

for a = 1, 3, 7, or 9 and k ≥ 6. This proves (7), and since we now know that thetrailing digit of pn cannot be 1, part (8) follows immediately.

With Proposition 4 in hand, we are almost ready to begin the search to 1045. How-ever, it is not so simple to compute pn for large numbers n. What is simple is comput-ing the inverse of the logarithmic integral function, li−1(n), and so we would like toknow how close this is to pn. The tools in the next section give us some guidance inthis regard.

4. BOUNDS. We will make use of the first Chebyshev function, θ(x) =∑p≤x

log p,

where p runs over prime numbers. We will also require li(x) =∫ x

0

dt

log t, the loga-

rithmic integral function. Here the “principal value” is taken at the singularity at t = 1;that is, if x > 1, then

li(x) = limy→0+

(∫ 1−y

0

dt

log t+

∫ x

1+y

dt

log t

).

This is the traditional way of defining li(x) and it has its advantages, but it admittedlymakes the function li(x) look very complicated, and doing so only adds a constant to



the perhaps more natural∫ x

2

dt

log t. The function li(x) is a much better approximation

to π(x) than is x/ log x and it is why we introduce it. In any event, li(x) is asymptoticto x/ log x as x→∞, in that

limx→∞

li(x)x/ log x

= 1.

(This can be easily proved using L’Hopital’s rule.) We shall also be using the inverseof li(x), namely li−1(x), which satisfies

limx→∞

li−1(x)x log x

= 1.

It is of interest to us because li−1(n) is a very good approximation to pn. How goodwe shall see shortly.

Let

A(x) = x− θ(x),B(x) = li(x)− π(x),

C(n) = pn − li−1(n).

We wish to find numerically explicit bounds for |C(n)|. Lemma 5 relates the functionsA and B, and Lemma 6 relates B and C. We will use these relations to ultimatelyobtain bounds for |C(n)|.

Lemma 5. For x > a > 2, we have

B(x)−B(a) =A(x)

log x− A(a)

log a+

∫ x

a

A(t)

t(log t)2dt.

Proof. This result follows from “partial summation,” a discrete analogue of integrationby parts. However, we may verify the identity directly. Note that∫ x

a

dt

t(log t)2=

1

log a− 1

log x,

so that ∫ x

a

θ(t)

t(log t)2dt =

∫ x

a

θ(a)

t(log t)2dt+

∑a<p≤x

∫ x

p

log p

t(log t)2dt

=θ(a)

log a− θ(a)

log x+∑

a<p≤x

(log p

log p− log p

log x

)

=θ(a)

log a− θ(x)

log x+ π(x)− π(a).

Since ∫ x

a

dt

(log t)2= li(x)− li(a)− x

log x+

a

log a,



we thus have∫ x

a

A(t)

t(log t)2dt =

∫ x

a

t− θ(t)t(log t)2

dt = B(x)−B(a)− A(x)

log x+A(a)

log a,

and the result is proved.

We will choose some convenient number for a where A(a), B(a) have been com-puted (a = 1019 in Proposition 10).

Lemma 6. For any integer n > 0, we have

|C(n)| ≤ |B(pn)| log(max{pn, li−1(n)}

).

Proof. We apply the mean value theorem to the function li on the interval with end-points pn and li−1(n) to obtain

li(pn)− n =pn − li−1(n)

log u,

for some value of u in the interval. Thus,

C(n) = B(pn) log u, (4)

and taking absolute values, the result follows.

We will split the positive integers into two intervals: those at most 1019 and thoseabove 1019. If we are in the lower range, then Buthe [2, Theorem 2] gives the followingstrong inequality. For 2 ≤ x ≤ 1019,

0 < B(x) <

√x

log x

(1.95 +

3.9

log x+

19.5

(log x)2

). (5)

This allows us to use (4) to obtain the following bound on C(n).

Lemma 7. For pn < 1019,

0 < C(n) <√pn

(1.95 +

3.9

log pn+

19.5

(log pn)2

).

Proof. This follows immediately from (4) and (5) provided log u ≤ log pn, where uis in the interval with endpoints pn and li−1(n). But (5) implies that li−1(n) < pnwhen pn < 1019.

Note that if we only know n and are not sure what pn is we can still use Lemma 7if we combine it with the simple upper bound from [7, (3.13)]:

pn < n(log n+ log log n), n ≥ 6. (6)

For example, we have (3) from Section 2. For another example, suppose n = 335.We compute that

li−1(335) = 2.05844182653518213541×1018,



with an error smaller than 0.01. The error bound given by (6) and Lemma 7 is lessthan 3×109. Thus pn has 19 digits and the leading 9 of them are 205844182. This pnis obviously not a Sheldon prime, as it will clearly fail the product property.

To complement our upper bound (6) we shall need the following lower bound forli−1(x).

Lemma 8. For x ≥ 12,218 we have

li−1(x) > x(log x+ log log x− 1).

Proof. This inequality is clearly equivalent (since li is an increasing function) to

li(x(log x+ log log x− 1)) < x

for x ≥ 12,218. Note that it holds at x = 12,218. So, it will follow if we show that

d

dx(li(x(log x+ log log x− 1)) < 1 (7)

in the same range. The derivative in (7) is

log x+ log log x+ 1/ log x

log x+ log(log x+ log log x− 1).

Letting z = log x, this derivative is

z + log z + 1/z

z + log(z + log z − 1)= 1− log(1 + (log z − 1)/z)− 1/z

z + log(z + log z − 1).

We thus would like to show this last numerator is positive. Using the inequalitylog(1 + w) > w/(1 + w) for w > 0, the numerator is larger than

log z − 1

z + log z − 1− 1

z.

It’s clear then that this is positive for large enough values of z, and we check in factthat z ≥ 8.46 is sufficient. This holds if x ≥ 5000, so we have shown (7) and thus thelemma.

For x ≥ 1019, we use the following estimate for |A(x)| from [5, Proposition 2.1]that uses bounds of Buthe [1].

Lemma 9. For x ≥ 1019, |A(x)| < εx, with ε = 2.3×10−8.

With Lemma 9, we can now construct our remaining upper bound for C(n).

Proposition 10. Let

E(x) =(5.5×109 + 2.3×10−8li(x) + 10−11x

)log x.

For pn > 1019, we have

|C(n)| < E(li−1(n)

).



Proof. Let a = 1019. Using (5), we have |B(a)| < 2×108. We now use Lemma 9 andddt(li(t)− t/ log t) = 1/(log t)2 to get for x > a,

|A(x)|log x

< 2.3×10−8 x

log x,

|A(a)|log a

< 5.3×109, and∫ x

a

|A(t)|t(log t)2

dt < 2.3×10−8∫ x

a

dt

(log t)2< 2.3×10−8

(li(x)− x

log x

).

We thus conclude from Lemma 5 that for x > 1019,

|B(x)| ≤ |B(a)|+ |A(a)|log a

+|A(x)|log x

+

∫ x

a

|A(t)|t(log t)2

dt

< 5.5×109 + 2.3×10−8li(x).

Let

E1(x) =(5.5×109 + 2.3×10−8li(x)

)log x,

so that from Lemma 6 we have for pn > 1019 that

|C(n)| ≤ E1

(max{pn, li−1(n)}

).

The proposition follows in the case that pn ≤ li−1(n). Suppose the reverse inequalityholds, that is, pn > li−1(n). We use the upper bound (that’s evidently an improvementon (6)!)

pn < n(log n+ log log n− 1 +

log log n− 2

log n

), n ≥ 688,383,

found in [4, Proposition 5.15]. With this and also using Lemma 8, we find that

E1(pn) < E(li−1(n))

for n > 1017, a range which includes pn > 1019 (using Lemma 7), so completing theproof.

Our calculations were performed using Mathematica. In particular, we usedthe built-in function LogIntegral[x] for li(x). Starting from the approximationx(log x + log log x − 1), we were then able to use a few iterations of Newton’smethod to compute li−1(x) for numbers x of interest to us.

5. PROOF OF THEOREM 1. We first search over any primes less than 1019. ByLemma 7, if pn < 1019 then n ≤ N := 2.341×1017. So we begin our search bycreating a list of all 7-smooth numbers up to N . This is quickly computed by creatinga list of numbers of the form 2a3b5c7d, with

0 ≤ a ≤ log2(N),

0 ≤ b ≤ log3(N/2a),



0 ≤ c ≤ log5(N/(2a3b)),

0 ≤ d ≤ log7(N/(2a3b5c)).

In particular, there are 57,776 integers of this form. We remove the 7,575 members ofthe table that are at most 1010 since we have previously searched over these numbersn to recover the 3 primes satisfying the product property.

We will use the properties of Proposition 4 to eliminate the remaining 50,201 valuesof n. First we remove those n where 100 | n or 54 | n, leaving 13,335 numbers. Forthese n, we compute li−1(n) and check via Lemma 7 that the leading digit is welldetermined. Similarly we check that the number of digits of pn is well determined.Then we select those with the leading digit of pn in {1, 3, 7, 9}. These filters reduceour list to a possible 6,893 candidates.

We then select those n where the top 5 digits of pn are given by the top 5 digits ofli−1(n). All but 68 values of n have this property. Using the same method as Propo-sition 2, we assume that all the remaining digits of pn are 9’s and check to see if theproduct of these 9’s and the top 5 known digits is at least n. If not, we can rule outn, and this eliminates all but 576 cases. For these cases, we check if the top 6 digitsare given by li−1(n), and all but 61 of them still have this property. We then repeatthe product test with the top 6 digits and this leaves only 180 numbers. Combined, ourthree remaining sets together total 309 = 68 + 61 + 180 possible candidates.

For these remaining numbers n, we compute rev(n). By part (8) of Proposition 4,we reduce to the 60 of them with first digit of prev(n) in {3, 7, 9}. Of these, 55 of themhave known top 5 digits. The 5 exceptions correspond to rev(n) being one of

4,019,155,056, 4,032,803,241, 4,079,545,092,

12,427,422,237, 29,794,252,274.

These are all small enough so that we can find the corresponding primes directly:

97,496,326,163, 97,841,660,857, 99,024,780,191,

316,109,730,941, 785,009,387,557.

They all have a digit 0 except for the first one, and that has an internal digit 1, and sothese 5 are ruled out by part (7) of Proposition 4.

With the remaining 55 numbers, we can again use the product test with rev(n),as described above, and this eliminates all but 6 of them. These are too large to findthe corresponding primes, but we can easily find how many digits the correspondingprimes have, and only 2 of the 6 have the same number of digits as the primes corre-sponding to n. For these two, we know the leading 6 digits of pn and the leading 5digits of prev(n), which would need to be the trailing 5 digits of pn if pn were indeed aSheldon prime. The product of these 11 digits times the appropriate power of 9 for thestill-unknown digits is too small for these to be Sheldon primes. This completes thesearch up to 1019.

For the remainder of the proof, we use Proposition 10. If pn < 1045, then n <9.746×1042. We compute the 7-smooths to this bound; there are 1,865,251 of them.Removing those less than 2.34×1017 and those divisible by 100 or 125 leaves a listof 213,449 remaining numbers. Each of these gives an unambiguous first digit for pn,and then selecting those where the first digit is in {1, 3, 7, 9} leaves 112,344. We thenverify that for each of these we can use li−1(n) to determine the exact number of digitsof pn.



We then test if the first 5 digits of pn are unambiguous and all but 168 of themhave this property. For those that do have the property, we multiply the top 5 digits byan appropriate power of 9 to get an upper bound on the product of the digits of pn,keeping only the 991 of them where this upper bound is at least n. We then repeat thisprocedure with the top 6 digits. All but 29 of them have the top 6 digits determined,and of the remaining values of n, all but 277 of them are discarded because the productof digits is too small. We then keep only those where the product of the first 6 digitsdivides n; there are 141 left.

We thus have a remaining set of size 338 = 168 + 29 + 141 numbers n. For these,we check that the number of digits and the first digit of prev(n) is determined fromli−1(rev(n)). We then discard those where the number of digits of pn is not equal tothe number of digits of prev(n) and those where the top digit of prev(n) is not in {3, 7, 9}.This leaves only 45 numbers. Each of these has li−1(rev(n)) able to determine the top5 digits of prev(n), and all of these values of n fail the test where we multiply the top 5digits of prev(n) and the appropriate power of 9 and check that against n.

This completes the proof that 73 is the unique Sheldon prime.

6. FUTURE WORK. Several generalizations and extensions of this concept natu-rally emerge from the above discussion. For instance, the product property of a Shel-don prime clearly rests on its base-10 representation. Can you classify all primes sat-isfying the product property in different bases? For instance, 226,697 is the 20,160thprime, and its base-9 representation is 3748659. Multiplying its base-9 digits togetherreturns 20,160 and so we can say 226,697 satisfies the product property in base-9.

Is there a meaningful way to describe a prime which nearly has the product prop-erty? For instance, p35 = 149. The product of the digits of 149 is 36, which is only 1away from 35, and hence 149 is quite close to having the product property.

For a positive integer n, let f(n) denote the product of the base-10 digits of pn.Then an index n for which pn has the product property satisfies f(n) = n, and con-versely. If we iterate the function f we can find some longer cycles. For example,f(1) = 2, f(2) = 3, f(3) = 5, f(5) = 1. Since a cycle must contain a number nsuch that f(n) ≥ n, it’s clear from Proposition 2 that there are only finitely manycycles. Can one find any others? Note that an iteration comes to an end as soon asa number n is encountered such that pn has a 0 digit. Otherwise an orbit eventuallyenters a cycle.

It is interesting to note that most primes do have a digit 0 in their decimal expansion,since the number of integers in [2, x] with no digit 0 is at most about x0.954 which issmall compared with π(x) (which we have seen is about x/ log x). One might guessthere are infinitely many primes missing the digit 0, and in fact, this was recentlyproved by Maynard [6].

We showed in Proposition 2 that there are at most finitely many primes with theproduct property. What about the mirror property? In addition to 73, the primes 2, 3, 5,7, and 11 all have the mirror property, which is easily verifed. They are all examples of“palindromic mirror primes” in that both pn and n are palindromes. A larger example,from [3], is

p8,114,118 = 143,787,341.

A heuristic argument suggests that there are infinitely many primes with the mirrorproperty, but the palindromic mirror primes occur more frequently than the mirrorproperty primes that are not palindromes. In fact, up to x there should be about

√log x

palindromic mirror primes, and about log log xmirror primes that are not palindromes.



ACKNOWLEDGMENTS. The episode of The Big Bang Theory that contained the discussion about 73, wasactually the 73rd of the series! The episode teleplay was written by Lee Aronsohn, Jim Reynolds, and MariaFerrari, and we thank them for bringing the special properties of the number 73 to a wide audience, and thatincluded us! We also thank David Saltzberg, who writes math on the whiteboards for the show, for his interestin this paper. We also thank Christian Axler for calling to our attention a small issue in the proof of Proposition10 and the referees for some useful comments.

REFERENCES

1. Buthe, J. (2016). Estimating π(x) and related functions under partial RH assumptions. Math. Comp. 85:2483–2498. doi.org/10.1090/mcom/3060

2. Buthe, J. (2018). An analytic method for bounding ψ(x). Math. Comp. 87: 1991–2009.doi.org/10.1090/mcom/3264

3. Byrnes, J., Spicer, C., Turnquist, A. (2015). The Sheldon conjecture. Math Horizons. 23(2): 12–15.doi.org/10.4169/mathhorizons.23.2.12

4. Dusart, P. (2018). Explicit estimates of some functions over primes. Ramanujan J. 45(1): 227–251.doi.org/10.1007/s11139-016-9839-4

5. Lichtman, J. D., Pomerance, C. (2018). Explicit estimates for the distribution of numbers free of largeprime factors. J. Number Theory. 183: 1–23. doi.org/10.1016/j.jnt.2017.08.039

6. Maynard, J. (2016). Primes with restricted digits. Available at: arxiv.org/abs/1604.010417. Rosser, J. B., Schoenfeld, L. (1962). Approximate formulas for some functions of prime numbers. Illinois

J. Math. 6: 64–94.

CARL POMERANCE is the John G. Kemeny Parents Professor of Mathematics Emeritus at Dartmouth Col-lege and is also Research Professor Emeritus at the University of Georgia. Former positions include middleschool teacher in Revere, MA, and member of the technical staff at Bell Labs. His research is principally inanalytic, combinatorial, and computational number theory. He considers Paul Erdos, who always appreciateda fun problem, his greatest influence.Department of Mathematics, Dartmouth College, Hanover NH [email protected]

CHRIS SPICER is an associate professor of Mathematics at Morningside College. He received his Ph.D.from North Dakota State University in 2010. He is an avid watcher of The Big Bang Theory and has alwaysbeen fascinated with math in popular culture.Department of Mathematical Sciences, Morningside College, Sioux City IA [email protected]


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